
If there is a loss of 40% when a good is sold at $\dfrac{2}{5}th$ of its earlier selling price. Find the profit% after selling the goods at an earlier selling price?
Answer
522.9k+ views
Hint: Start by taking the selling price as x for both the conditions stated in the question. Try to form the relation between the two selling prices as stated in the question and apply the loss(%) formula in order to get one of the unknown values and use the relation to find the other one . Apply the profit(%) formula to get the final desired value.
Complete step-by-step answer:
Given, loss of 40% at selling price$\dfrac{2}{5}th$ of earlies selling price.
Let , the first selling price be =${x_1}$
Then , second selling price be =${x_2}$
$\therefore {x_2} = \dfrac{2}{5}{x_1} \to (1)$
We know , ${\text{Loss}}(\% ) = \dfrac{{{\text{Cost price - selling price}}}}{{{\text{Cost price}}}} \times 100$
$
40 = \dfrac{{{\text{C}}{\text{.P}}{\text{. - }}{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \times 100 \\
\Rightarrow \dfrac{{40}}{{100}} = \dfrac{{{\text{C}}{\text{.P}}{\text{. - }}{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \\
\Rightarrow 0.4 = 1 - \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \\
\Rightarrow \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} = 1 - 0.4 \\
\Rightarrow \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} = 0.6 \\
\Rightarrow {x_2} = 0.6 \times {\text{C}}{\text{.P}}{\text{.}} \\
$
Now , substituting this value in relation 1, we get
$
{x_2} = \dfrac{2}{5}{x_1} \\
0.6 \times {\text{C}}{\text{.P}}{\text{. = 0}}{\text{.4}} \times {x_1} \\
\Rightarrow {x_1} = 1.5 \times {\text{C}}{\text{.P}}{\text{.}} \\
\\
$
Now, we know ${\text{Profit}}(\% ) = \dfrac{{{\text{selling price - Cost price}}}}{{{\text{Cost price}}}} \times 100$
Profit earned while selling at first selling price,
$
{\text{Profit}}(\% ) = \dfrac{{{\text{1}}{\text{.5}} \times {\text{C}}{\text{.P}}{\text{. - C}}{\text{.P}}{\text{.}}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \times 100 \\
\Rightarrow {\text{Profit}}(\% ) = (1.5 - 1) \times 100 \\
\Rightarrow {\text{Profit}}(\% ) = (0.5) \times 100 \\
\therefore {\text{Profit}}(\% ) = 50\% \\
$
Therefore , the profit earned is 50% when sold at an earlier price.
Note: All the formulas related to profit and loss must be well known to the students . Attention must be given while substituting the values in the equation and also while forming the required relations.
Always remember if you get profit as negative value , then it is a loss and not a profit.
Complete step-by-step answer:
Given, loss of 40% at selling price$\dfrac{2}{5}th$ of earlies selling price.
Let , the first selling price be =${x_1}$
Then , second selling price be =${x_2}$
$\therefore {x_2} = \dfrac{2}{5}{x_1} \to (1)$
We know , ${\text{Loss}}(\% ) = \dfrac{{{\text{Cost price - selling price}}}}{{{\text{Cost price}}}} \times 100$
$
40 = \dfrac{{{\text{C}}{\text{.P}}{\text{. - }}{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \times 100 \\
\Rightarrow \dfrac{{40}}{{100}} = \dfrac{{{\text{C}}{\text{.P}}{\text{. - }}{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \\
\Rightarrow 0.4 = 1 - \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \\
\Rightarrow \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} = 1 - 0.4 \\
\Rightarrow \dfrac{{{x_2}}}{{{\text{C}}{\text{.P}}{\text{.}}}} = 0.6 \\
\Rightarrow {x_2} = 0.6 \times {\text{C}}{\text{.P}}{\text{.}} \\
$
Now , substituting this value in relation 1, we get
$
{x_2} = \dfrac{2}{5}{x_1} \\
0.6 \times {\text{C}}{\text{.P}}{\text{. = 0}}{\text{.4}} \times {x_1} \\
\Rightarrow {x_1} = 1.5 \times {\text{C}}{\text{.P}}{\text{.}} \\
\\
$
Now, we know ${\text{Profit}}(\% ) = \dfrac{{{\text{selling price - Cost price}}}}{{{\text{Cost price}}}} \times 100$
Profit earned while selling at first selling price,
$
{\text{Profit}}(\% ) = \dfrac{{{\text{1}}{\text{.5}} \times {\text{C}}{\text{.P}}{\text{. - C}}{\text{.P}}{\text{.}}}}{{{\text{C}}{\text{.P}}{\text{.}}}} \times 100 \\
\Rightarrow {\text{Profit}}(\% ) = (1.5 - 1) \times 100 \\
\Rightarrow {\text{Profit}}(\% ) = (0.5) \times 100 \\
\therefore {\text{Profit}}(\% ) = 50\% \\
$
Therefore , the profit earned is 50% when sold at an earlier price.
Note: All the formulas related to profit and loss must be well known to the students . Attention must be given while substituting the values in the equation and also while forming the required relations.
Always remember if you get profit as negative value , then it is a loss and not a profit.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Trending doubts
The aviation fuel used in the engines of jet airplanes class 10 physics CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE

What is the full form of POSCO class 10 social science CBSE

Draw a labelled diagram of the human digestive system class 10 biology CBSE

What is potential and actual resources
